awk search on a variable

dazdaz · 04-13-2012, 08:34 PM

Hi,

I would like to do the following, however /matchvar/ is the string being searched and not the variable passed to gawk from the shell, that is being searched for.

Do I have to use some special awk quoting, or a different method of searching ?

Code:

 $ match=1000 ; echo '2000 1000' | gawk -F' ' -v matchvar=$match '{if ($2 ~ /matchvar/){print "it worked"}}'

rknichols · 04-13-2012, 10:57 PM

Quote:

Originally Posted by dazdaz

Code:

 $ match=1000 ; echo '2000 1000' | gawk -F' ' -v matchvar=$match '{if ($2 ~ /matchvar/){print "it worked"}}'

With those slashes you are trying to match against the literal characters they enclose, i.e., the 8-character string "matchvar". Just use the name alone.

Code:

 gawk -F' ' -v matchvar=$match '{if ($2 ~ matchvar){print "it worked"}}'

firstfire · 04-14-2012, 12:18 AM

Hi.

From man awk:

Quote:

In the special case that FS is a single space, fields are
separated by runs of spaces and/or tabs and/or newlines.

AFAIK, space is the default value of FS, so you can omit -F' '. You can also perform matching in the pattern section of the script:

Code:

$ match=1000 ; echo '2000 1000' | gawk -v matchvar=$match '$2 ~ matchvar {print "it worked"}'
it worked

dazdaz · 04-14-2012, 05:36 PM

Hi, I also need to match 0, and from further testing, if I set match to 10, I also see an invalid positive match result.

Afaik ~ is for strings, so it worked when I used == for numbers instead.

Code:

$ match=0 ; echo '2000 1000' | gawk -F' ' -v matchvar=$match '{if ($2 ~ matchvar){print "it worked"}}'
it worked

$ match=10 ; echo '2000 1000' | gawk -F' ' -v matchvar=$match '{if ($2 ~ matchvar){print "it worked"}}'
it worked

* This works :

Code:

$ match=0 ; echo '2000 1000' | gawk -F' ' -v matchvar=$match '{if ($2 == matchvar){print "it worked"}}'

grail · 04-15-2012, 06:21 AM

If this is SOLVED please mark it as so.